Question

If $T=\sum _{r=1}^{18}\frac{(-1{)}^{r-1}\cdot r\cdot {}^{18}{C}_r}{(r+1)}$ then $\left(\frac{1}{T}-10\right)$ is equal to

Answer 1

Answer: 9

$\frac{r}{(r+1)}\cdot {}^{18}{C}_r=\left(1-\frac{1}{r+1}\right)\cdot {}^{18}{C}_r$
$={}^{18}{C}_r-\frac{\lfloor 18}{(r+1)\lfloor r\lfloor 18-r}$
$={}^{18}{C}_r-\frac{1}{19}{}^{19}{C}_{r+1}$
$\therefore T=\sum _{r=1}^{18}(-1{)}^{r-1}\cdot \frac{r}{(r+1)}\cdot {}^{18}{C}_r$
$={}^{18}{C}_1-{}^{18}{C}_2+{}^{18}{C}_3-\dots$
$={}^{18}{C}_0-\frac{1}{19}\left(-{}^{19}{C}_0+{}^{19}{C}_1\right)$
$=1+\frac{1}{19}-1=\frac{1}{19}$
$\Rightarrow \frac{1}{T}=19$